course 2 1-11 addition and subtraction equations 1-11 addition and subtraction equations course 2...

Course 2

1-11 Addition and Subtraction Equations1-11 Addition and Subtraction Equations

Course 2

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Course 2

1-11 Addition and Subtraction Equations

Warm UpDetermine if the given numbers are solutions to the given equations.

1. x = 2 for 4x = 9

2. x = 5 for 8x + 2 = 42

3. x = 4 for 3(x – 2) = 10

no

yes

no

Course 2


Problem of the Day

Four couples have dinner together. The wives are Ginny, Helen, Sarah, and Bridget. The husbands are Mark, Alex, Stephen, and Henry. Who is married to whom?

• Sarah is Mark’s sister.

• Sarah introduced Henry to his wife.

• Bridget has 2 brothers, but her husband is an only child.

• Ginny is married to Stephen.

Ginny and Stephen, Helen and Mark, Sarah andAlex, Bridget and Henry

Course 2


Learn to solve one-step equations by using addition or subtraction.

Course 2


Vocabulary

Addition Property of Equalityinverse operationsSubtraction Property of Equality

Course 2


To solve an equation means to find a solution to the equation. To do this, isolate the variable—that is, get the variable alone on one side of the equal sign.

x = 8 – 5 x + 5 = 87 – 3 = y 7 = 3 + y

The variables are isolated. The variables are not isolated.

Recall that an equation is like a balanced scale. If you increase or decrease the weights by the same amount on both sides, the scale will remain balanced.

Course 2


You can add the same amount to both sides of an equation, and the statement will still be true.

2 + 3 = 5+ 4 + 4

2 + 7 = 9

ADDITION PROPERTY OF EQUALITYADDITION PROPERTY OF EQUALITY

Words Numbers Algebra

x = y+ z = + z

x + z = y+ z

Course 2


Use inverse operations when isolating a variable. Addition and subtraction are inverse operations, which means that they “undo” each other.

Course 2


Solve the equation b – 7 = 24. Check your answer.

Additional Example 1: Solving an Equation by Addition

b – 7 = 24+ 7 +7

b = 31

Think: 7 is subtracted from b, soadd 7 to both sides to isolate b.

Check

b – 7 = 24

31 – 7 = 24?

24 = 24?

Substitute 31 for b.

31 is a solution.

Course 2


Check It Out: Example 1

Solve the equation y – 3 = 21. Check your answer.

y – 3 = 21+ 3 +3

y = 24

Think: 3 is subtracted from y, soadd 3 to both sides to isolate y.

Check

y – 3 = 21

24 – 3 = 21?

21 = 21?

Substitute 24 for y.

24 is a solution.

Course 2


You can subtract the same amount from both sides of an equation, and the statement will still be true.

4 + 7 = 11 –3 – 3

4 + 4 = 8

SUBTRACTION PROPERTY OF EQUALITYSUBTRACTION PROPERTY OF EQUALITY

Words Numbers Algebra

x = y– z = – z

x – z = y – z

Course 2


Solve the equation t + 14 = 29. Check your answer.

Additional Example 2: Solving an Equation by Subtraction

t + 14 = 29– 14 – 14

t = 15

Checkt + 14 = 29

15 + 14 = 29?

29 = 29?

Think: 14 is added to t, so subtract 14 from both sides to isolate t.

Substitute 15 for t.

15 is a solution.

Course 2



Solve the equation x + 11 = 36. Check your answer.

x + 11 = 36– 11 – 11

x = 25

Check

Think: 11 is added to x, so subtract 11 from both sides to isolate x.

x + 11 = 36

25 + 11 = 36?

36 = 36?

Substitute 25 for x.

25 is a solution.

Course 2


The Giants scored 13 points in a game against Dallas. They scored 7 points for a touchdown and the rest of their points for field goals. How many points did they score on field goals?

Additional Example 3: Sports Application

Let f represent the field goal points.

7 points + field goal points = points scored 7 + f = 13

7 + f = 13– 7 – 7

f = 6They scored 6 points on field goals.

Subtract 7 from both sides to isolate f.

Course 2



A basketball player scored 23 points during a game. Of those points, 3 were from 3-point goals and the remainder were 2 point goals. How many points did he score with 2 point goals?

Let x equal the points scored by 2 point goals.

3 point goals + 2 point goals = points scored

3 + x = 23

3 + x = 23– 3 – 3

x = 20

He scored 20 points from 2 point goals.

Subtract 3 from both sides to isolate x.

Course 2


Lesson Quiz

Solve each equation. Check your answer.

1. x – 9 = 4

2. y + 6 = 72

3. 21 = n – 41

4. 127 = w + 31

5. 81 = x – 102

y = 66; 66 + 6 = 72

x = 13; 13 – 9 = 4

n = 62; 21 = 62 – 41

w = 96; 127 = 96 + 31

x = 183; 81 = 183 – 1026. Tamika has sold 16 dozen cookies this week.

This was 7 dozen more than she sold last week. Write and solve an equation to find how many dozen cookies she sold last week.

x + 7 = 16; 9 dozen

Course 2

1-11 Addition and Subtraction Equations1-12 Multiplication and Division Equations

Course 2

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Course 2


Warm UpSolve.

1. x + 5 = 9

2. x – 34 = 72

3. 124 = x – 39

x = 4

x = 106

x = 163

Course 2


Problem of the Day

What 4-digit number am I?• I am greater than 4,000 and less than 5,000.• The sum of my hundreds digit and my ones digit is 9.• Twice my tens number is 2 more than my thousands digit.• The product of my hundreds digit and my ones digit is 0.

• I am not an even number. 4,039

Course 2


Learn to solve one-step equations by using multiplication or division.

Course 2


Vocabulary

Multiplication Property of Equality

Division Property of Equality

Course 2


Like addition and subtraction, multiplicationand division are inverse operations. They “undo” each other.

•

÷

Course 2


If a variable is divided by a number, you can often use multiplication to isolate the variable. Multiply both sides of the equation by the number.

Course 2


Solve the equation = 13. Check your answer.

Additional Example 1: Solving an Equation by Multiplication

h2

h2 = 13

h2 = 13(2)(2)

h = 26Check

h2 = 13

13 = 13?

Think: h is divided by 2, so multiply both sides by 2 to isolate h.

Substitute 26 for h.

26 is a solution.

262

?= 13

Course 2



Solve the equation = 30. Check your answer.x5

x5

= 30

x5 = 30(5)(5)

x = 150Check

x5 = 30

Think: x is divided by 5, so multiply both sides by 5 to isolate x.

30 = 30?

Substitute 150 for x.

150 is a solution.

150 5

? = 30

Course 2


You cannot divide by 0.

Remember!

Course 2


If a variable is multiplied by a number, you can often use division to isolate the variable. Divide both sides of the equation by the number.

Course 2


Solve the equation 51 = 17x. Check your answer.

Additional Example 2: Solving an Equation by Division

51 = 17x

51 = 17x17 17

3 = x

Check51 = 17x

Think: x is multiplied by 17,so divide both sides by 17 to isolate x.

Substitute 3 for x.

3 is a solution.

51 = 17(3)?

51 = 51?

Course 2



Solve the equation 76 = 19y. Check your answer.

76 = 19y

76 = 19y19 19

4 = y

Check76 = 19y

Think: y is multiplied by 19,so divide both sides by 19 to isolate y.

Substitute 4 for y.

4 is a solution.

76 = 19(4)?

76 = 76?

Course 2


Trevor’s heart rate is 78 beats per minute. How many times does his heart beat in 10 seconds?

Additional Example 3: Health Application

Use the given information to write an equation, where b is the number of heart beats in 10 seconds.

If you count your heart beats for 10 seconds andthen multiply that by 6, you can find your heart ratein beats per minute.

Course 2


Beats in 10s times 6 = beats per minutes

Additional Example 3 Continued

b · 6 = 78

6b = 78

6b = 786 6b = 13

Trevor’s heart beats 13 times 10 seconds.

Think: b is multipliedby 6, so divide bothsides by 6 to isolate b.

Course 2



During a stock car race, one driver is able to complete 68 laps in 1 hour. How many laps would he finish in 15 minutes?

If you count the number of laps in 15 minutes and multiply by 4, you can find the number of laps completed in 1 hour.

Use the given information to write an equation, where n is the number of laps completed in 15 minutes.

Course 2


Check It Out: Example 3 Continued

Laps in 15 min times 4 = Laps in 1 hour

n · 4 = 68

4n = 68

4n = 684 4n = 17

The driver would complete 17 laps in 15 minutes.

Think: L is multipliedby 4, so divide bothsides by 4 to isolate n.

Course 2


Lesson Quiz: Part I

Solve the equation. Check your answer.

1. 12 = 4x

2. 18z = 90

3. 12 =

4. 840 = 12y

5.

z = 5; 18 5 = 90

x = 3; 12 = 4 3

y = 70; 840 = 12 70

x4

h22

= 9

x = 48; 12 = x4

h = 198; = 919822

Course 2


Lesson Quiz: Part II

6. The cost of each ticket at the carnival was $0.25.

Li bought $7.50 worth of tickets. How many

tickets did she buy?

30

course 2 1-11 addition and subtraction equations 1-11 addition and subtraction equations course 2...

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